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Contests
National and Regional Contests
Turkey Contests
Turkey Team Selection Test
2016 Turkey Team Selection Test
2016 Turkey Team Selection Test
Part of
Turkey Team Selection Test
Subcontests
(8)
9
1
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Polynomials with Neccessary Equal Degrees
p
p
p
is a prime. Let
K
p
K_p
K
p
be the set of all polynomials with coefficients from the set
{
0
,
1
,
…
,
p
−
1
}
\{0,1,\dots ,p-1\}
{
0
,
1
,
…
,
p
−
1
}
and degree less than
p
p
p
. Assume that for all pairs of polynomials
P
,
Q
∈
K
p
P,Q\in K_p
P
,
Q
∈
K
p
such that
P
(
Q
(
n
)
)
≡
n
(
m
o
d
p
)
P(Q(n))\equiv n\pmod p
P
(
Q
(
n
))
≡
n
(
mod
p
)
for all integers
n
n
n
, the degrees of
P
P
P
and
Q
Q
Q
are equal. Determine all primes
p
p
p
with this condition.
8
1
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Obtuse-Angled n-gon
All angles of the convex
n
n
n
-gon
A
1
A
2
…
A
n
A_1A_2\dots A_n
A
1
A
2
…
A
n
are obtuse, where
n
≥
5
n\ge5
n
≥
5
. For all
1
≤
i
≤
n
1\le i\le n
1
≤
i
≤
n
,
O
i
O_i
O
i
is the circumcenter of triangle
A
i
−
1
A
i
A
i
+
1
A_{i-1}A_iA_{i+1}
A
i
−
1
A
i
A
i
+
1
(where
A
0
=
A
n
A_0=A_n
A
0
=
A
n
and
A
n
+
1
=
A
1
A_{n+1}=A_1
A
n
+
1
=
A
1
). Prove that the closed path
O
1
O
2
…
O
n
O_1O_2\dots O_n
O
1
O
2
…
O
n
doesn't form a convex
n
n
n
-gon.
7
1
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Arithmetic Sequence in Every Intersection
A
1
,
A
2
,
…
A
k
A_1, A_2,\dots A_k
A
1
,
A
2
,
…
A
k
are different subsets of the set
{
1
,
2
,
…
,
2016
}
\{1,2,\dots ,2016\}
{
1
,
2
,
…
,
2016
}
. If
A
i
∩
A
j
A_i\cap A_j
A
i
∩
A
j
forms an arithmetic sequence for all
1
≤
i
<
j
≤
k
1\le i <j\le k
1
≤
i
<
j
≤
k
, what is the maximum value of
k
k
k
?
6
1
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Concyclic Points in an Isosceles Triangle
In a triangle
A
B
C
ABC
A
BC
with
A
B
=
A
C
AB=AC
A
B
=
A
C
, let
D
D
D
be the midpoint of
[
B
C
]
[BC]
[
BC
]
. A line passing through
D
D
D
intersects
A
B
AB
A
B
at
K
K
K
,
A
C
AC
A
C
at
L
L
L
. A point
E
E
E
on
[
B
C
]
[BC]
[
BC
]
different from
D
D
D
, and a point
P
P
P
on
A
E
AE
A
E
is taken such that
∠
K
P
L
=
9
0
∘
−
1
2
∠
K
A
L
\angle KPL=90^\circ-\frac{1}{2}\angle KAL
∠
K
P
L
=
9
0
∘
−
2
1
∠
K
A
L
and
E
E
E
lies between
A
A
A
and
P
P
P
. The circumcircle of triangle
P
D
E
PDE
P
D
E
intersects
P
K
PK
P
K
at point
X
X
X
,
P
L
PL
P
L
at point
Y
Y
Y
for the second time. Lines
D
X
DX
D
X
and
A
B
AB
A
B
intersect at
M
M
M
, and lines
D
Y
DY
D
Y
and
A
C
AC
A
C
intersect at
N
N
N
. Prove that the points
P
,
M
,
A
,
N
P,M,A,N
P
,
M
,
A
,
N
are concyclic.
4
1
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Polynomial Having Values of a Sequence
A sequence of real numbers
a
0
,
a
1
,
…
a_0, a_1, \dots
a
0
,
a
1
,
…
satisfies the condition
∑
n
=
0
m
a
n
⋅
(
−
1
)
n
⋅
(
m
n
)
=
0
\sum\limits_{n=0}^{m}a_n\cdot(-1)^n\cdot\dbinom{m}{n}=0
n
=
0
∑
m
a
n
⋅
(
−
1
)
n
⋅
(
n
m
)
=
0
for all large enough positive integers
m
m
m
. Prove that there exists a polynomial
P
P
P
such that
a
n
=
P
(
n
)
a_n=P(n)
a
n
=
P
(
n
)
for all
n
≥
0
n\ge0
n
≥
0
.
2
1
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Movie Collections of Students
In a class with
23
23
23
students, each pair of students have watched a movie together. Let the set of movies watched by a student be his movie collection. If every student has watched every movie at most once, at least how many different movie collections can these students have?
1
1
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Property of Orthocenter
In an acute triangle
A
B
C
ABC
A
BC
, a point
P
P
P
is taken on the
A
A
A
-altitude. Lines
B
P
BP
BP
and
C
P
CP
CP
intersect the sides
A
C
AC
A
C
and
A
B
AB
A
B
at points
D
D
D
and
E
E
E
, respectively. Tangents drawn from points
D
D
D
and
E
E
E
to the circumcircle of triangle
B
P
C
BPC
BPC
are tangent to it at points
K
K
K
and
L
L
L
, respectively, which are in the interior of triangle
A
B
C
ABC
A
BC
. Line
K
D
KD
KD
intersects the circumcircle of triangle
A
K
C
AKC
A
K
C
at point
M
M
M
for the second time, and line
L
E
LE
L
E
intersects the circumcircle of triangle
A
L
B
ALB
A
L
B
at point
N
N
N
for the second time. Prove that
K
D
M
D
=
L
E
N
E
⟺
Point P is the orthocenter of triangle ABC
\frac{KD}{MD}=\frac{LE}{NE} \iff \text{Point P is the orthocenter of triangle ABC}
M
D
KD
=
NE
L
E
⟺
Point P is the orthocenter of triangle ABC
5
1
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Functional Number Theory (Turkey IMO TST 2016 P5)
Find all functions
f
:
N
→
N
f: \mathbb{N} \to \mathbb{N}
f
:
N
→
N
such that for all
m
,
n
∈
N
m,n \in \mathbb{N}
m
,
n
∈
N
holds
f
(
m
n
)
=
f
(
m
)
f
(
n
)
f(mn)=f(m)f(n)
f
(
mn
)
=
f
(
m
)
f
(
n
)
and
m
+
n
∣
f
(
m
)
+
f
(
n
)
m+n \mid f(m)+f(n)
m
+
n
∣
f
(
m
)
+
f
(
n
)
.